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We prove that x*Px < or eq. x*Qx, for all x in C^n (C : complex numbers) if and only if x*Q^-1 x < or eq. x*P^-1 x for all x in C^n.

I guess I should use the definition of a hermitian positive definite matrix being

x*Px > 0 , for all x in C^n but I am not sure how to proceed to get both the P and Q in the inequality.

Should I try and multiply both sides of the inequality by x and x*?